Friction

Push a heavy crate across a floor. Nothing happens. Push harder: still nothing. Push harder again and suddenly it breaks loose and slides more easily than it took to start it. Somewhere in there the floor was pushing back on the crate with exactly the force you were applying, and then it stopped being able to.

That "exactly the force you were applying" is the whole lesson. In the previous lesson you drew free body diagrams and every force on them had a known size: weight was $mg$, the normal force was whatever the surface pressed with. Friction is the first force you will meet whose size is not handed to you. Most of the time you have to solve for it.

Two laws, two grammars

Friction acts along the contact surface and opposes relative sliding. It comes in two kinds, and the difference between them is the difference between an equation and an inequality.

Once two surfaces are already sliding across each other, the friction force has a fixed magnitude set by how hard they are pressed together:

$f_k = \mu_k N$

Here $N$ is the normal force, the perpendicular push of the surface on the object, and $\mu_k$ is the coefficient of kinetic friction: a dimensionless number, usually between 0.05 and 1, that depends on what the two materials are. Notice what is not in that formula: the speed, the area of contact, the weight. Only $N$ and the materials. A brick lying flat and the same brick standing on end feel the same kinetic friction.

Static friction, when the surfaces are not sliding, is a different animal. It supplies whatever force is needed to prevent sliding, up to a limit:

$f_s \le \mu_s N$

Read that as a budget. Push the crate with 20 N and it stays put: static friction is 20 N, not $\mu_s N$. Push with 50 N and it stays put: static friction is 50 N. The number $\mu_s N$ is the largest cheque the surface can write. Only at the instant the crate is about to slip does the inequality become an equality, and that is the only moment you are allowed to write $f_s = \mu_s N$.

Worked example 1: a car that locks its brakes

A car travelling at $20\ \mathrm{m/s}$ locks its wheels on a level road with $\mu_k = 0.70$. How far does it skid?

The wheels are locked, so the tyres slide: kinetic friction, an equation, not a budget. On level ground the only vertical forces are weight and normal force, so $N = mg$. The friction force is $f_k = \mu_k mg$ and it is the only horizontal force, so Newton's second law along the road gives

$ma = -\mu_k mg \quad\Longrightarrow\quad a = -\mu_k g = -(0.70)(9.8) = -6.86\ \mathrm{m/s^2}.$

The mass cancelled. That is worth pausing on: a loaded truck and an empty one, same tyres, skid the same distance. Now use the kinematic relation with no time in it, $v^2 = v_0^2 + 2a d$, with final speed zero:

$d = \frac{v_0^2}{2\mu_k g} = \frac{20^2}{2(0.70)(9.8)} = \frac{400}{13.72} = 29.2\ \mathrm{m}.$

The same three lines run backwards give you $\mu_k$ from a measured skid, or from a measured deceleration: if a block slows at $2.5\ \mathrm{m/s^2}$ on a level table, then $\mu_k = a/g = 2.5/9.8 = 0.255$.

Slopes and the angle of repose

Tilt a plank with a block on it and raise the angle $\theta$ slowly. Set up axes along the slope and perpendicular to it, rather than horizontal and vertical: that way the acceleration, if there is any, lies along one axis. Gravity $mg$ points straight down, which splits into $mg\sin\theta$ down the slope and $mg\cos\theta$ into it. Perpendicular to the surface nothing accelerates, so

$N = mg\cos\theta.$

Along the slope, gravity demands $mg\sin\theta$ of friction to hold the block. The budget is $\mu_s N = \mu_s mg\cos\theta$. The block sits still as long as demand does not exceed budget:

$mg\sin\theta \le \mu_s\, mg\cos\theta \quad\Longrightarrow\quad \tan\theta \le \mu_s.$

The mass cancelled again, and so did $g$. The critical angle, called the angle of repose, is $\theta_{\max} = \arctan\mu_s$. Tilt a ramp until a block just starts to slide at 32°, and you have measured $\mu_s = \tan 32^\circ = 0.625$. Go the other way: a carton with $\mu_s = 0.60$ holds up to $\arctan 0.60 = 31.0°$.

Past that angle the block slides, friction switches to $\mu_k N$ pointing up the slope, and

$a = g(\sin\theta - \mu_k\cos\theta).$

For a 20° slope with $\mu_k = 0.25$: $a = 9.8(0.342 - 0.25 \times 0.940) = 9.8(0.342 - 0.235) = 1.05\ \mathrm{m/s^2}$.

One sign trap hides here. If the block is launched up the slope, it is still moving, so friction still has magnitude $\mu_k mg\cos\theta$, but it now points down the slope, because it opposes the motion, not gravity. The two effects add:

$a = g(\sin\theta + \mu_k\cos\theta).$

A 6.0 kg block sent up a 25° incline with $\mu_k = 0.30$ decelerates at $9.8(0.423 + 0.30 \times 0.906) = 9.8(0.423 + 0.272) = 6.81\ \mathrm{m/s^2}$, much harder than it would accelerate coming back down.

The friction cone

There is a picture that makes all of this one idea instead of several. Add the normal force and the friction force as vectors and call the result the contact force. Since $f \le \mu_s N$, the contact force can lean away from the perpendicular by at most $\arctan \mu_s$. Sweep that limit around and you sweep out a cone.

The rule is then: if the other forces on the object point into the cone, the surface can hold it; if they point outside, nothing can. The angle of repose is the moment gravity leaves the cone. It also explains why shoving a crate downward at a steep angle never helps you start it moving: pushing at an angle outside the cone adds to the driving component and to $N$ in exactly the ratio that keeps you outside it.

Worked example 2: the best angle to pull

A 25 kg sled is dragged at constant speed over ground with $\mu_k = 0.50$. You may choose the rope angle. What is the smallest tension that works?

Blank paper first: draw the sled, the weight $mg$ down, the normal force $N$ up, friction $f_k$ backwards, and the rope tension $T$ at angle $\theta$ above the horizontal. Constant speed means zero acceleration in both directions.

Vertically, the rope now carries part of the load, so the ground carries less:

$N + T\sin\theta = mg \quad\Longrightarrow\quad N = mg - T\sin\theta.$

This is the step people miss. Tilting the rope lightens the sled, which shrinks $N$, which shrinks friction. Horizontally,

$T\cos\theta = \mu_k N = \mu_k(mg - T\sin\theta).$

Collect the $T$ terms: $T\cos\theta + \mu_k T\sin\theta = \mu_k mg$, so

$T(\theta) = \frac{\mu_k mg}{\cos\theta + \mu_k\sin\theta}.$

Check it on an easier case first. A 20 kg trunk, $\mu_k = 0.30$, rope at 30°: $T = (0.30)(20)(9.8) / (0.866 + 0.30 \times 0.500) = 58.8/1.016 = 57.9$ N.

To minimise $T$ you maximise the denominator. Write $\cos\theta + \mu\sin\theta = \sqrt{1+\mu^2}\,\cos(\theta - \arctan\mu)$ (that is the standard trick for turning $a\cos\theta + b\sin\theta$ into a single cosine) and the largest it can be is $\sqrt{1+\mu^2}$, reached when $\theta = \arctan\mu$. So the best rope angle is the friction cone angle, and

$T_{\min} = \frac{\mu_k mg}{\sqrt{1+\mu_k^2}} = \frac{(0.50)(25)(9.8)}{\sqrt{1.25}} = \frac{122.5}{1.118} = 110\ \mathrm{N}.$

Pulling flat would have cost 122.5 N. The curve is flat near its minimum, which is why nobody has to be precise about it in real life.

The same bookkeeping runs in reverse when a force presses a block into a slope. Push horizontally with force $F$ on a 10 kg block on a 40° incline with $\mu_s = 0.30$, and now $F$ has a component into the surface (raising $N$) and a component up the slope. Perpendicular: $N = mg\cos\theta + F\sin\theta$. Along the slope, on the verge of sliding down, friction acts up at full budget: $F\cos\theta + \mu_s N = mg\sin\theta$. Solving for the minimum $F$ gives 42.2 N. The structure is identical: resolve, note that $N$ now depends on the unknown, then solve.

Two bodies, two budgets

Stack a block on a block, or tie two together, and each contact gets its own $N$, its own $\mu$, and its own budget. The system moves as one rigid object only while every contact stays inside its budget.

The procedure is always the same three steps. Assume nothing slips and find the common acceleration by treating the whole stack as one object. Then isolate one body and ask how much friction that acceleration requires at each contact. Finally compare each required value with that contact's own $\mu_s N$.

Worked example 3. A 2.0 kg block sits on a 5.0 kg block on a frictionless floor, with $\mu_s = \mu_k = 0.40$ between them. A horizontal 40 N force is applied to the lower block. Find the lower block's acceleration.

Assume they move together. Total mass 7.0 kg, one external horizontal force:

$a = \frac{40}{7.0} = 5.71\ \mathrm{m/s^2}.$

Now isolate the top block. The only horizontal force on it is friction from the block underneath, so that friction alone must produce $5.71\ \mathrm{m/s^2}$:

$f_{\text{needed}} = (2.0)(5.71) = 11.4\ \mathrm{N}.$

The budget at that contact is $\mu_s N = (0.40)(2.0)(9.8) = 7.84$ N. Demand beats budget, so the assumption fails: the top block slides backwards relative to the bottom one. Redo it with kinetic friction fixed at 7.84 N and separate accelerations. The top block gets $a_1 = 7.84/2.0 = 3.92\ \mathrm{m/s^2}$. The bottom block feels the 40 N push forward and the top block's friction dragging backward on it (Newton's third law):

$a_2 = \frac{40 - 7.84}{5.0} = \frac{32.16}{5.0} = 6.43\ \mathrm{m/s^2}.$

Run the same machinery to find the largest force that keeps a stack together. For 3.0 kg on 7.0 kg with $\mu_s = 0.40$ between them and a frictionless floor, the top block can be accelerated at most at $\mu_s g = 3.92\ \mathrm{m/s^2}$, so the whole 10 kg can be pushed with at most $F = (10)(3.92) = 39.2$ N.

When the floor is rough too, there are two contacts to check and the loser is not always the obvious one. Put a 2.0 kg block on a 4.0 kg block on the ground, $\mu_s = 0.30$ between them and $0.20$ underneath, and push the top block. The top block slips once $F > (0.30)(2.0)(9.8) = 5.88$ N. The ground contact carries all 6.0 kg, so it does not give until $F > (0.20)(6.0)(9.8) = 11.8$ N. The upper surface fails first at 5.88 N, even though it is the one with the larger coefficient.

That last check is a real problem type. A 2.0 kg block with $\mu_k = 0.10$ is tied ahead of a 3.0 kg block with $\mu_k = 0.40$ on a 30° slope. Treat the pair as one body: the effective coefficient is the weighted average $(0.10 \times 2.0 + 0.40 \times 3.0)/5.0 = 0.28$, so $a = 9.8(0.500 - 0.28 \times 0.866) = 2.52\ \mathrm{m/s^2}$. Then isolate the front block, where the string pulls backward: $T = m_1 g\sin\theta - \mu_1 m_1 g\cos\theta - m_1 a = 9.80 - 1.70 - 5.05 = 3.06$ N.

When the whole thing is accelerating

Friction problems get one level harder when the surface itself accelerates. A block rides on the 20° face of a wedge, and the wedge is driven horizontally at $A = 3.0\ \mathrm{m/s^2}$ into the hill. What is the smallest $\mu_s$ that keeps the block from sliding on the face?

Stay in the ground frame and use axes along and across the slope. The block's acceleration is horizontal with magnitude $A$, so its component along the slope is $A\cos\theta$ and across the slope is $A\sin\theta$. Perpendicular to the face, with the block pressed in,

$N = mg\cos\theta + mA\sin\theta.$

Along the face, gravity pulls down-slope at $mg\sin\theta$, the required acceleration accounts for $mA\cos\theta$ up-slope, and friction $f$ makes up the difference:

$mg\sin\theta - f = -mA\cos\theta \ \Longrightarrow\ f = m(g\sin\theta - A\cos\theta).$

Divide by $N$, and the mass drops out once more:

Driving the wedge into the hill did most of the holding, so barely any friction is needed. Set $A=0$ and you recover $\tan\theta$, the angle-of-repose answer, which is a good check that the algebra is right.

Multi-stage motion is the other common escalation: a block slides 3.0 m down a 30° ramp with $\mu_k = 0.20$, then onto a floor with $\mu_k = 0.35$. Do it in two pieces, carrying the speed across the join. On the ramp, $a = 9.8(0.500 - 0.20 \times 0.866) = 3.20\ \mathrm{m/s^2}$, so $v^2 = 2(3.20)(3.0) = 19.2\ \mathrm{m^2/s^2}$. On the floor, $a = -\mu_k g = -3.43\ \mathrm{m/s^2}$, so $d = 19.2/(2 \times 3.43) = 2.80$ m. You never need $v$ itself, only $v^2$.

How to recognise a friction problem

The word "friction" is usually there, but the useful signal is finer than that.

If the problem says "minimum force", "maximum force", "just barely", "on the verge", or "about to slip", you are at the threshold and you may write $f_s = \mu_s N$. That phrase is the permission slip for the equals sign.

If the object is already sliding, you have $f_k = \mu_k N$ immediately and the only work left is getting the direction right: opposite the velocity, not opposite the applied force. Blocks moving up a rough slope are where this bites.

If the object is at rest and the problem asks for the friction force, do not use either coefficient until you have checked. Solve the equilibrium equations for $f$, then compare with $\mu_s N$ to confirm it is affordable.

If there is more than one contact surface, expect to assume no slipping, solve for a common acceleration, and check every contact against its own budget. Whichever fails first is where the motion begins.

And if only an angle appears with no masses at all, the answer is almost certainly $\tan\theta = \mu$ in some disguise: the angle of repose, the best rope angle, or the ladder condition $\tan\theta \ge 1/(2\mu_s)$, which is the same friction budget argument run through a torque balance instead of a force balance. For $\mu_s = 0.40$ that gives $\arctan(1.25) = 51.3°$.

Next you will tie these blocks together properly and send the string over pulleys, where the constraint that the rope has a fixed length does the work that the no-slip assumption did here: Tension, Pulleys & Connected Systems.